For finding the solution of (1), we may seek a function yyy which is product of two functions:

y⁢(x)=u⁢(x)⁢v⁢(x)yxuxvx\displaystyle y(x)\;=\;u(x)v(x)

(2)

One of these two can be chosen freely; the other is determined according to (1).

We substitute (2) and the derivatived⁢yd⁢x=u⁢d⁢vd⁢x+v⁢d⁢ud⁢xdydxudvdxvdudx\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx} in (1), getting
u⁢d⁢vd⁢x+v⁢d⁢ud⁢x+P⁢u⁢v=QudvdxvdudxPuvQu\frac{dv}{dx}+v\frac{du}{dx}+Puv=Q, or